Harmonic Number LightOJ

In mathematics, the nth harmonic number is the sum of the reciprocals of the first n natural numbers。

Hn = 1 + 1 / 2 +１/3 +…… + 1/n

In this problem, you are given n, you have to find Hn.

Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 108).

Output
For each case, print the case number and the nth harmonic number. Errors less than 10-8 will be ignored.

Sample Input
12
1
2
3
4
5
6
7
8
9
90000000
99999999
100000000
Sample Output
Case 1: 1
Case 2: 1.5
Case 3: 1.8333333333
Case 4: 2.0833333333
Case 5: 2.2833333333
Case 6: 2.450
Case 7: 2.5928571429
Case 8: 2.7178571429
Case 9: 2.8289682540
Case 10: 18.8925358988
Case 11: 18.9978964039
Case 12: 18.9978964139

暴力打表，这道题学会了按区域来保存的技巧，（把结果分成很多段，保存每段的第一个结果，搜索时找到对应区间再暴力该区间就可以）

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<queue>#include<algorithm>#define ll long long#define inf 0x3f3f3f3f#define maxn 1e8+7using namespace std;double a[2000100];void Init(){    a[0] = 0.0;    a[1] = 1.0;    double ans = 1;    for( int i = 2; i < maxn; i ++)    {        ans += 1.0 / i;        if( i % 50 == 0)            a[i/50] = ans;    }   }int main(){    int t,n;    Init();//  for(int i = 0;i <= 10; i++)//      printf("%lf\n",a[i]);    scanf("%d",&t);    for(int i = 1; i <= t;i ++)     {        scanf("%d",&n);        int tmp = n / 50;        double ans = a[tmp];        for(int j = tmp * 50 + 1; j <= n; j ++)        {            ans += 1.0 / j;        }        printf("Case %d: %.9lf\n",i,ans);    }    return 0;}